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Homework Help: One Sided Limits

  1. Sep 6, 2008 #1
    The question is: Find

    lim x-->2+ (x+3) * |x+2| / x+2

    I am confused about the absolute value problems. I cannot seem to grasp in my mind how to define this problem piecewised. Any help to get me started would be appreciated.
     
  2. jcsd
  3. Sep 6, 2008 #2

    HallsofIvy

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    Science Advisor

    Actually, there is no problem with the absolute value in this problem! |x| "breaks" when x= 0 so |x+2| "breaks" when x+ 2= 0 or x= -2. Since here you are taking the limit as x goes to 2, not -2, for x "close to 2" (larger than -2) x+2> 0 and this is exactly like
    [tex]\lim_{x\rightarrow 2} (x+3)*(x+2)/(x+2)= \lim_{x\rightarrow 2} x+ 3= 2+ 3= 5[/tex].

    Now, if the limit were being taken as x goes to -2, THEN you would need to worry about "one sided" limits. No matter how close x is to -2, it could still be either < -2 or > -2.
    If x< -2, then x+ 2 is negative and so |x+2|= -(x+2). |x+2|/(x+2)= -(x+2)/(x+2)= -1 so
    (x+3)|x+2|/(x+2)= -(x+3) and the limit, as x goes to -2 from below, is -(-2+3)= -1.

    But if x> -2, then x+2>0 so |x+2|= x+2 and |x+2|/(x+2)= 1 so (x+3)|x+2|/(x+2)= x+ 3 and the limit, as x goes to -2 from above, is (-2+3)= 1.
     
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